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For a 4000kg car, the rated power of the engine is 40kW. After the car is started, the speed is 0.5m/s square In line motion, the resistance in motion is always 2000N. Find 1. The time for the car to do uniform acceleration. 2. Find the maximum speed that the car can reach
V(max) = P/f = 40000w/2000N = 20m/s
t = V(max)/a = 20m/s / 0.5m/s^2 = 40s
The rated power of automobile engine with mass of M = 4T kg is 40 kW, and the automobile starts to move uniformly in a straight line with acceleration of a = 0.5 m / S2 from standstill, and the resistance constant is f = 2000 n. calculate the power P of the engine when the automobile speed is 5 m / S
Driving force F = F + Ma = 2000 + 0.5 * 4000 = 4000N
P=Fv=5*4000=20kW
The rated power of automobile engine is 30kW and the mass is 2000kg. When the automobile runs on the horizontal road, the resistance is 0.1 times of the vehicle weight. (1) the maximum speed that the automobile can reach on the road. (2) how long can the process last if the automobile keeps the acceleration of 1m / S2 for uniform linear acceleration since it is stationary. (3) the acceleration when the automobile speed is 10m / S?
(1) According to P = FVM, the maximum speed is: VM = pf = 300002000m / S = 15m / S. (2) according to Newton's second law: f-f = ma, the solution is: F = F + Ma = 2000 + 2000 × 1n = 4000N, according to P = FV, the final speed of uniform acceleration is obtained
The rated power of automobile engine is 60kW, and the mass of automobile is 5000kg. When the automobile runs on the horizontal road, the resistance is 0.1 times of the vehicle weight, g = 10m / s
G = 10m / S ^ 2. (1) after the vehicle starts from standstill with the rated power unchanged, what is the speed when the acceleration of the vehicle is 2m / S ^ 2?
It should be 48m / s. I'm just a high school student. Do you want to go through the process?
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